00:01
Okay, we're supposed to solve this partial differential equation using the method of characteristics with this is our initial value or our initial function.
00:12
So the first step is to write out the equations for the characteristics, which we get from examining this equation.
00:22
The dots represent derivatives along the characteristic curves.
00:26
And if we compare the second and third equations, we get this.
00:37
And we can solve that just by integrating.
00:44
Alpha is some constant, something that's constant along the characteristic curves.
00:53
So we get x is alpha over u or ux is equal to alpha.
00:58
So that's one of our options.
01:03
Okay to get the second one here's what i'm going to do i'm going to take this equation i'm going to solve it for x and that becomes x is minus x dot over 2y and i'm going to plug it back into that y equation so i get y dot is minus x dot over 2y and then i can multiply by 2y.
01:43
All right.
01:44
And then i can integrate that.
01:47
I get y squared is minus x plus a constant.
01:54
Let's call it beta.
01:59
And that tells me that beta is y squared minus or plus x.
02:08
Right.
02:10
So that means my general solution is an arbitrary function f of alpha, and beta, which is some function f that depends on ux and y squared plus x...